Question

How do you differentiate `f(x)= sin(x-2)^3` using the chain rule.?

Answer 1

`f'(x)=3(x-2)^2cos[(x-2)^3]`

Explanation:

The chain rule states that if `y=f(u) and u=f(x)` then
`dy/dx=(dy)/(du)*(du)/dx`

Furthermore, the power rule states that
`d/dx[u(x)]^n=n*[u(x)]^(n-1)*(du)/dx`

And then finally we also have that `d/dxsin[u(x)]=cosu*(du)/dx`.

So application of these rules yields the following result :

`d/dx[sin(x-2)^3]=cos[(x-2)^3]*3(x-2)^2*1`

`=3(x-2)^2cos[(x-2)^3]`